solution conflict!


by gulsen
Tags: conflict, solution
gulsen
gulsen is offline
#1
Jan4-06, 12:44 PM
P: 224
I have a basic differential equation:
[tex]\frac{dy}{dx} = x + y, y(0) = 1[/tex]

Now, when I try to solve this by making it exact
[tex]\mu \frac{dy}{dx} + \mu y = \mu x[/tex]
I get [tex]\mu = e^{-x}[/tex] and solution [tex]-x-1[/tex]. This doesn't satisfy the initial condition. But when I try to solve it as a non-homogenous equation as:
[tex]\frac{dy}{dx} + y= x[/tex]
I get
[tex]y_p = 2e^x, y_c = -x-1[/tex]
so
[tex]y = 2e^x-x-1[/tex]

Which seems to be a correct & full solution. What was I missing in the first try?
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Eratosthenes
Eratosthenes is offline
#2
Jan4-06, 01:18 PM
P: 74
You are solving dy/dx + y = x which is wrong since the original equation is
dy/dx = x + y, or dy/dx - y = x so now u(x) = e^(-x) and so on and it works out.


And yes, the answer is y = 2e^x - x - 1
D H
D H is online now
#3
Jan4-06, 01:54 PM
Mentor
P: 14,427
Quote Quote by gulsen
I have a basic differential equation:
[tex]\frac{dy}{dx} = x + y, y(0) = 1[/tex]
Now, when I try to solve this by making it exact
[tex]\mu \frac{dy}{dx} + \mu y = \mu x[/tex]
I assume this was just a typo, but that equation should be
[tex]\mu \frac{dy}{dx} - \mu y = \mu x[/tex]
I get [tex]\mu = e^{-x}[/tex] and solution [tex]-x-1[/tex]. This doesn't satisfy the initial condition.
Your error: you forgot the constant of integration. You should have obtained
[tex]y = -(x+1) + c/\mu = ce^x - (1+x)[/tex]
and then solved for the initial condition [itex]y(0)=1[/itex] yielding [itex]c=2[/itex] or
[tex]y = 2e^x -(x+1)[/tex]
But when I try to solve it as a non-homogenous equation as:
[tex]\frac{dy}{dx} + y= x[/tex]
I get
[tex]y_p = 2e^x, y_c = -x-1[/tex]
so
[tex]y = 2e^x-x-1[/tex]
You're nomenclature is backward here. The solution to the homogeneous equation is called the complementary function and is denoted as [itex]y_c[/itex]. The complementary function generally involves arbitrary constants. A solution to the inhomogeneous equation is called a particular function and is denoted as [itex]y_p[/itex].
In this case, the solution to the homogeneous equation [itex]y^\prime-y=0[/itex] is
[tex]y_c = ce^x[/tex]
where [itex]c[/itex] is an arbitrary constant and
[tex]y_p = -(x+1)[/tex]
is a particular solution to the inhomogeneous equation. Combining these,
[tex]y = ce^x - (1+x)[/tex]
which meets the initial conditions when [itex]c=2[/itex].

gulsen
gulsen is offline
#4
Jan19-06, 11:19 PM
P: 224

solution conflict!


Oh, that was just a typo. Thanks D_H, it ends up that I've forgotten the integration constant.
About naming, take it easy, I'm just a freshman!


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